# Compound and double angle identities

> IB Mathematics: Analysis and Approaches HL · Geometry & Trigonometry
> Source: https://www.owlsprep.com/study/ib-math-aa-hl-u3-compound-and-double-angle-identities/

This subtopic covers core compound angle and double angle trigonometric identities. You will learn to derive these identities, apply them to simplify expressions, prove more complex identities, and solve trigonometric equations — all common exam question types.

**Prerequisites:** [Basic trigonometric and Pythagorean identities](https://www.owlsprep.com/study/ib-math-aa-hl-u3-basic-trigonometric-identities/); [Unit circle and exact trigonometric values](https://www.owlsprep.com/study/ib-math-aa-hl-u3-unit-circle-trigonometry/)

## Learning objectives

- Derive compound and double angle trigonometric identities
- Apply identities to simplify trigonometric expressions
- Prove complex trigonometric identities using these formulas
- Solve trigonometric equations using compound/double angle identities

## Compound Angle Identities

**Compound Angle Identity** — A trigonometric identity that expresses a function of a sum or difference of two angles in terms of functions of the individual angles

*Example:* $\cos(A-B)$ written in terms of $\cos A, \cos B, \sin A, \sin B$

The core compound angle identities for sine and cosine are the foundation for all other identities in this subtopic:

$$\cos(A + B) = \cos A \cos B - \sin A \sin B$$

$$\cos(A - B) = \cos A \cos B + \sin A \sin B$$

$$\sin(A + B) = \sin A \cos B + \cos A \sin B$$

$$\sin(A - B) = \sin A \cos B - \cos A \sin B$$

For tangent, divide $\sin(A \pm B)$ by $\cos(A \pm B)$ to get:

$$\tan(A \pm B) = \frac{\tan A \pm \tan B}{1 \mp \tan A \tan B}$$

**Worked example:** Find the exact value of $\cos 15^\circ$.

1. Write $15^\circ$ as the difference of two angles with known exact values: $15^\circ = 45^\circ - 30^\circ$.
2. Apply the compound angle identity for cosine of a difference:
3. $$\cos(45^\circ - 30^\circ) = \cos 45^\circ \cos 30^\circ + \sin 45^\circ \sin 30^\circ$$
4. Substitute known exact values:
5. $$= \left(\frac{\sqrt{2}}{2}\right)\left(\frac{\sqrt{3}}{2}\right) + \left(\frac{\sqrt{2}}{2}\right)\left(\frac{1}{2}\right)$$
6. Simplify to get the final result:
7. $$= \frac{\sqrt{6} + \sqrt{2}}{4}$$

> **Exam tip:** Remember the sign rule: cosine flips the sign, sine keeps the sign for the compound term.

## Double Angle Identities

**Double Angle Identity** — A special case of the compound angle identity where $A = B$, giving a function of $2A$ in terms of functions of $A$

Derive double angle identities by setting $B = A$ in the compound angle formulas. For cosine, we get three equivalent forms using the Pythagorean identity $\sin^2 A + \cos^2 A = 1$:

$$\sin 2A = 2\sin A \cos A$$

$$\begin{aligned} \cos 2A &= \cos^2 A - \sin^2 A \\ &= 2\cos^2 A - 1 \\ &= 1 - 2\sin^2 A \end{aligned}$$

$$\tan 2A = \frac{2\tan A}{1 - \tan^2 A}$$

> **tip**
>
> Memorize all three forms of $\cos 2A$. Choosing the right form will drastically simplify problems, eliminating unnecessary algebra.

**Worked example:** Simplify $4\sin \theta \cos \theta$ to a single multiple-angle trigonometric term.

1. Factor out 2 from the expression:
2. $$4\sin \theta \cos \theta = 2 \left(2\sin \theta \cos \theta\right)$$
3. Recognize $2\sin \theta \cos \theta = \sin 2\theta$ from the double angle identity for sine:
4. $$2(\sin 2\theta) = 2\sin 2\theta$$

## Proving Trigonometric Identities

Proving identities is a very common IB exam question. The standard approach is to start with the more complicated side of the identity and apply identities to simplify it until it matches the other side.

> **tip**
>
> Never move terms across the equals sign when proving identities in IB exams. You must work on only one side to transform it into the other side to get full marks.

**Worked example:** Prove that $\frac{1 - \cos 2\theta}{\sin 2\theta} = \tan \theta$.

1. Start with the left-hand side (LHS). Choose the double angle form of $\cos 2\theta = 1 - 2\sin^2 \theta$ to match the $1 - \cos 2\theta$ term:
2. $$\text{LHS} = \frac{1 - (1 - 2\sin^2 \theta)}{\sin 2\theta}$$
3. Simplify the numerator, then substitute $\sin 2\theta = 2\sin \theta \cos \theta$:
4. $$= \frac{2\sin^2 \theta}{2\sin \theta \cos \theta}$$
5. Cancel common factors ($2\sin \theta$, for $\sin \theta \neq 0$ where the identity is defined):
6. $$= \frac{\sin \theta}{\cos \theta} = \tan \theta = \text{Right-hand side (RHS)}$$
7. The identity is proven.

**Check your understanding**

Test your understanding of which $\cos 2A$ form to choose:

1. Which form of $\cos 2\theta$ is optimal to prove $\frac{1 - \cos 2\theta}{\sin 2\theta} = \tan \theta$?

   - A. $\cos^2 \theta - \sin^2 \theta$
   - B. $2\cos^2 \theta - 1$
   - C. $1 - 2\sin^2 \theta$

   *Why:* Correct! This form cancels the 1 in the numerator, leaving a simple $\sin^2 \theta$ term that simplifies perfectly with the denominator.

## Solving Trigonometric Equations

Most trigonometric equations in exams have multiple angles (e.g. $\theta$ and $2\theta$). We use double angle identities to rewrite the entire equation in terms of a single angle, then solve as a polynomial or standard trigonometric equation.

**Worked example:** Solve $2\sin 2\theta = \tan \theta$ for $0 \leq \theta \leq \pi$.

1. Substitute $\sin 2\theta = 2\sin \theta \cos \theta$ and $\tan \theta = \frac{\sin \theta}{\cos \theta}$:
2. $$2(2\sin \theta \cos \theta) = \frac{\sin \theta}{\cos \theta}$$
3. Multiply through by $\cos \theta$ (for $\cos \theta \neq 0$) and rearrange:
4. $$4\sin \theta \cos^2 \theta - \sin \theta = 0$$
5. Factor out the common term $\sin \theta$:
6. $$\sin \theta (4\cos^2 \theta - 1) = 0$$
7. Set each factor equal to zero and solve in the interval: $\sin \theta = 0$ gives $\theta = 0, \pi$. $4\cos^2 \theta -1 =0$ gives $\cos \theta = \pm \frac{1}{2}$, giving $\theta = \frac{\pi}{3}, \frac{2\pi}{3}$.
8. Check that $\cos \theta = 0$ gives $\theta = \frac{\pi}{2}$, which makes $\tan \theta$ undefined, so it is not a solution.
9. The full solution set is:
10. $$\theta = 0, \frac{\pi}{3}, \frac{2\pi}{3}, \pi$$

*Calculator:* forbidden

## Common pitfalls

- **Wrong:** Memorizing the wrong sign for $\cos(A+B)$, writing $\cos A \cos B + \sin A \sin B$.
  - Why it fails: The sign for cosine compound angles is opposite to the angle operation.
  - Correct: Use the mnemonic: cos flips the sign, sine keeps the sign. So $A+B$ gives a minus for cosine, plus for sine.
- **Wrong:** Canceling $\sin \theta$ from both sides of an equation, leading to lost solutions.
  - Why it fails: Canceling removes the case where $\sin \theta = 0$, which is a valid solution.
  - Correct: Move all terms to one side and factor out common terms, then set each factor equal to zero to find all solutions.
- **Wrong:** Incorrectly assuming $\sin 2A = 2\sin A$.
  - Why it fails: The 2 cannot be factored out of the trigonometric function.
  - Correct: Remember the correct identity: $\sin 2A = 2\sin A \cos A$, which is not equal to $2\sin A$ for most values of $A$.
- **Wrong:** Using the wrong form of $\cos 2A$, leading to unnecessary complicated algebra.
  - Why it fails: Not matching the form of $\cos 2A$ to the existing terms in the expression.
  - Correct: Choose the form of $\cos 2A$ that will cancel constants or simplify with existing terms, e.g. use $1-2\sin^2 A$ when you have a $1 - \cos 2A$ term.
- **Wrong:** Forgetting to exclude points that make denominators zero when proving identities.
  - Why it fails: Identities are only valid for values where both sides are defined.
  - Correct: Always note that the identity holds for all values where both sides are defined, and exclude any points that make denominators zero.

## Cheatsheet

| Identity Type | Core Formulas |
| --- | --- |
| Compound Angle | $\cos(A \pm B) = \cos A \cos B \mp \sin A \sin B$ |
|  | $\sin(A \pm B) = \sin A \cos B \pm \cos A \sin B$ |
|  | $\tan(A \pm B) = \frac{\tan A \pm \tan B}{1 \mp \tan A \tan B}$ |
| Double Angle | $\sin 2A = 2\sin A \cos A$ |
|  | $\cos 2A = \cos^2 A - \sin^2 A = 2\cos^2 A - 1 = 1 - 2\sin^2 A$ |
|  | $\tan 2A = \frac{2\tan A}{1 - \tan^2 A}$ |

## What's next

Compound and double angle identities are the foundation for all advanced trigonometry in IB AA HL, appearing in topics from integration to differential equations. Mastery of these identities is critical to solving almost every complex trigonometry problem you will encounter in the exam. Next, you will extend these identities to other forms, and apply them to more complex problem types.

- [2D Coordinate Geometry](https://www.owlsprep.com/study/ib-math-aa-hl-u3-2d-coordinate-geometry/)
- [3D coordinate geometry](https://www.owlsprep.com/study/ib-math-aa-hl-u3-3d-coordinate-geometry/)
- [Vector fundamentals and dot product](https://www.owlsprep.com/study/ib-math-aa-hl-u3-vector-fundamentals-and-dot-product/)

---

From [OwlsPrep](https://www.owlsprep.com) — free study guides for A-Level, IB, AP and IGCSE, written against the official syllabus. Canonical page: https://www.owlsprep.com/study/ib-math-aa-hl-u3-compound-and-double-angle-identities/